Prove Isomorphism of Rotation Function in R^2

[beetle2]

Homework Statement

The problem is as follows:



Let f : R^2 map to R^2 be rotation through an angle of theta radians about the origin.

Prove that f is an isomorphism.

Homework Equations

Let f :R^2 \rightarrow R^2

The Attempt at a Solution

I know that the rotation can be expressed as the 2 x 2 matrix


cos(theta) -sin(theta)

Sin(theta) cos(theta)


And its inverse I believe is


cos(theta) -sin(theta)

-sin(theta) cos(theta)



Do I first show that f :R^2 \rightarrow R^2 is a linear transformation
by closure of addition and scaler multiplication by using x,y elements of R² and some scaler k


Say let the 2x2 matrix:

cos(theta) -sin(theta)

Sin(theta) cos(theta) = A


We need to show


A(x+y) = A(x) + A(y)


cos(theta)x -sin(theta)y = cos(theta)x + -sin(theta)y = cos(theta)x -sin(theta)y

sin(theta)x cos(theta)y sin(theta)x + cos(theta)y sin(theta)x cos(theta)y


likewise

A(kx+ky) = K A(x+y)

cos(theta)kx + -sin(theta)ky = cos(theta)xk -sin(theta)yk

sin(theta)kx + cos(theta)ky sin(theta)xk cos(theta)yk


Do I need to use the inverse or can I use an assumption some how?


Basically I’m having trouble with knowing how to prove that the function is 1-1 and surgective.


As well as constructing the proof logically and stating the proof clearly.



If anyone can help that would be great.



Regards

 

[VeeEight]

Try writing it out as in f(a,b)=...for (a,b) in R^2

To show that it is an isomorphism you should show that the function is 1-1, onto, and that it is a homomorphism. You mentioned that it may be a linear transformation. Did you check that?

 

[beetle2]

Yes It is a linear Transformation.

Let
<br /> \[ \left( \begin{array}{ccc}<br /> cos(\theta) &amp; -sin(\theta) \\<br /> sin(\theta) &amp; cos(\theta) \end{array} \right)\] = A.\\<br /> <br />
Than
<br /> <br /> A \times\[ \left( \begin{array}{ccc}<br /> x \\<br /> 0 \end{array} \right)\] + A \times\[ \left( \begin{array}{ccc}<br /> 0 \\<br /> y \end{array} \right)\] <br /> = \[ \left( \begin{array}{ccc}<br /> cos(\theta)x &amp; -sin(\theta)y \\<br /> sin(\theta)x &amp; cos(\theta)y \end{array} \right)\]<br /> <br /> <br />

Sorry I'm not to good a latex yet.




I Think the inverse might be:

<br /> \[ \left( \begin{array}{ccc}<br /> cos(\theta)x &amp; -sin(\theta)y \\<br /> -sin(\theta)x &amp; cos(\theta)y \end{array} \right)\]<br /> <br />
For the homomorphism the transformation is to R² as well so that is shown.

As of how to show it is 1-1 I'm don't know how. I'm not sure wether I need to use the inverse function or just use the assumption to show for f:G\rightarrow W there must be a function<br /> <br /> &lt;br /&gt; f:W\rightarrow G = f(x)^{-1}= w(f(x)) <br /> x is an element of G<br /> <br /> regards

 

[Dick]

I think the inverse of [[cos(theta),-sin(theta)],[sin(theta),cos(theta)]] might be [[cos(theta),sin(theta)],[-sin(theta),cos(theta)]]. (Each bracketed pair is a row of the matrix - I'm not that hot at latex either). Can you check that? If a linear transformation R^2->R^2 has an inverse, then it's 1-1.

 

[boneill3]

I Did the following on my calculator


A \times\[ \left( \begin{array}{ccc}x \\0 \end{array} \right)\] + A \times\[ \left( \begin{array}{ccc}0 \\y \end{array} \right)\] = \[ \left( \begin{array}{ccc}cos(\theta)x &amp; -sin(\theta)y \\sin(\theta)x &amp; cos(\theta)y \end{array} \right)\]

I then multiplied the result by the inverse as you said:

\[ \left( \begin{array}{ccc}cos(\theta)x &amp; sin(\theta)y \\-sin(\theta)x &amp; cos(\theta)y \end{array} \right)\]

which gave me

<br /> <br /> \[ \left( \begin{array}{ccc}x &amp; \\y \end{array} \right)\]

Is that right ?

As both functions have dim(2) have I done enough to show that the function is isomorphic ?

 

[HallsofIvy]

Well, yes, that is the whole point of a "inverse", isn't it- to "undo" the original operation. But you didn't really need the "x y" matrix:
\begin{bmatrix}cos(\theta) &amp; -sin(\theta) \\ sin(\theta) &amp; cos(\theta)\end{bmatrix}\begin{bmatrix}cos(\theta) &amp; sin(\theta) \\ -sin(\theta) &amp; cos(\theta)\end{bmatrix}= \begin{bmatrix} 1 &amp; 0 \\ 0 &amp; 1\end{bmatrix}.

Of course the opposite of "rotate through angle \theta" is "rotate through angle -\theta". So the inverse of
\begin{bmatrix}cos(\theta) &amp; -sin(\theta) \\ sin(\theta) &amp; cos(\theta)\end{bmatrix}
is
\begin{bmatrix}cos(-\theta) &amp; -sin(-\theta) \\ sin(-\theta) &amp; cos(-\theta)\end{bmatrix}= \begin{bmatrix}cos(\theta) &amp; sin(\theta) \\ -sin(\theta) &amp; cos(\theta)\end{bmatrix}
since "sine" is an odd function and "cosine" is an even function.

 

[boneill3]

Thanks for all your help guys.